Codeforces Round #806 (Div. 4)

Codeforces Round #806 (Div. 4)

A

Problem - A - Codeforces

挨个判断字符是不是等于大小写的"YES"

// Problem: A. YES or YES?
// Contest: Codeforces - Codeforces Round #806 (Div. 4)
// URL: https://codeforces.com/contest/1703/problem/A
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
char a[3];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        cin >> a;
        if ((a[0] == 'Y' || a[0] == 'y') && (a[1] == 'e' || a[1] == 'E') && (a[2] == 'S' || a[2] == 's'))
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }
    return 0;
}

B

Problem - B - Codeforces

模拟，如果第一次出现则多加一次

// Problem: B. ICPC Balloons
// Contest: Codeforces - Codeforces Round #806 (Div. 4)
// URL: https://codeforces.com/contest/1703/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
int a[26];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        memset(a, 0, sizeof a);
        int n;
        cin >> n;
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            char x;

            cin >> x;
            a[x - 'A']++;
            sum++;
            if (a[x - 'A'] == 1)
                sum++;
        }
        cout << sum << endl;
    }
    return 0;
}

C

Problem - C - Codeforces

模拟，$+10\%10$即可解决$0\to9$或$9\to0$的变化

// Problem: C. Cypher
// Contest: Codeforces - Codeforces Round #806 (Div. 4)
// URL: https://codeforces.com/contest/1703/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
int a[110];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        for (int i = 0; i < n; i++)
            cin >> a[i];
        for (int i = 0; i < n; i++)
        {
            int k;
            cin >> k;
            int res = a[i];
            while (k--)
            {
                char x;
                cin >> x;
                if (x == 'D')
                {
                    res = (res + 11) % 10;
                }
                else
                {
                    res = (res + 9) % 10;
                }
            }
            cout << res << " ";
        }
        cout << endl;
    }
    return 0;
}

D

Problem - D - Codeforces

模拟，对每个字符串的各个位置进行判断

// Problem: D. Double Strings
// Contest: Codeforces - Codeforces Round #806 (Div. 4)
// URL: https://codeforces.com/contest/1703/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        vector<string> v;
        map<string, int> m;
        for (int i = 0; i < n; i++)
        {
            string s;
            cin >> s;
            v.push_back(s);
            m[s]++;
        }
        for (int i = 0; i < n; i++)
        {
            bool flag = 0;
            for (int j = 1; j < v[i].size(); j++)
            {
                string s1 = v[i].substr(0, j);
                string s2 = v[i].substr(j);
                if (m.count(s1) && m.count(s2))
                    flag = 1;
                if (flag)
                    break;
            }
            if (flag)
                cout << 1;
            else
                cout << 0;
        }
        cout << endl;
    }
    return 0;
}

E

Problem - E - Codeforces

只有01，那么每个位置反转后的3个位置如果0多，则将1变为0，如果1多，则将0变为1

// Problem: E. Mirror Grid
// Contest: Codeforces - Codeforces Round #806 (Div. 4)
// URL: https://codeforces.com/contest/1703/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
char c[110][110];
int a[110][110];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
            {
                cin >> c[i][j];
                a[i][j] = c[i][j] - '0';
            }

        int sum = 0;
        for (int i = 0; i < (n + 1) / 2; i++)
        {
            for (int j = 0; j < n / 2; j++)
            {
                int temp = a[i][j];
                temp = temp + (a[j][n - 1 - i] + a[n - 1 - i][n - 1 - j] + a[n - 1 - j][i]);
                sum += min(temp, 4 - temp);
                // cout << sum << endl;
            }
        }
        cout << sum << endl;
    }
    return 0;
}

F

Problem - F - Codeforces

将所有$a_i<i$的数存入新的数组v，取数对$\{v_i,v_j\}$满足$i<j$

// Problem: F. Yet Another Problem About Pairs Satisfying an Inequality
// Contest: Codeforces - Codeforces Round #806 (Div. 4)
// URL: https://codeforces.com/contest/1703/problem/F
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>

using namespace std;
const int N = 2e5 + 10;
int a[N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int tt;
    cin >> tt;
    for (int i = 1; i <= tt; i++)
    {
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
            cin >> a[i];
        }
        long long res = 0;
        vector<int> v;
        for (int i = 1; i <= n; i++)
        {
            if (a[i] >= i)
            {
                continue;
            }
            res += (long long)(lower_bound(v.begin(), v.end(), a[i]) - v.begin());
            v.push_back(i);
        }
        cout << res << '\n';
    }
}

G

Problem - G - Codeforces

一定是先用完好钥匙再用坏钥匙

$\log_{2}(10^9)\approx30$

所以用了坏钥匙之后的30个箱子之后硬币会变成0

dp

// Problem: G. Good Key, Bad Key
// Contest: Codeforces - Codeforces Round #806 (Div. 4)
// URL: https://codeforces.com/contest/1703/problem/G
// Memory Limit: 256 MB
// Time Limit: 3000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n, k;
        cin >> n >> k;
        int a[n];
        for (int i = 0; i < n; i++)
        {
            cin >> a[i];
        }
        long long ans = 0;
        long long sum = 0;
        for (int i = -1; i < n; i++)
        {
            long long now = sum;
            for (int j = i + 1; j < min(n, i + 32); j++)
            {
                int copy = a[j];
                copy >>= j - i;
                now += copy;
            }
            ans = max(ans, now);
            if (i + 1 != n)
            {
                sum += a[i + 1] - k;
            }
        }
        cout << ans << endl;
    }
    return 0;
}